5+3 б 1.sin10+cos20 2.sin20+cos10 3.cos20-sin10 4.cos10-sin20

Question

Алгебра: 5+3 б 1.sin10+cos20 2.sin20+cos10 3.cos20-sin10 4.cos10-sin20

shtylev2018 · Accepted Answer

1) а^2 + 2а - 3 = ( а - 1 )( a + 3 )
D = 4 + 12 = 16 = 4^2
a1 = ( - 2 + 4 ) : 2 = 1
a2 = ( - 2 - 4 ) : 2 = - 3
2) 2a/( a + 3 ) + 1/( а - 1 ) - 4/( ( а - 1 )( а + 3 )) = ( 2а( a - 1 ) + a + 3 - 4 ) / ( ( a - 1 )( a + 3 )) = ( 2а^2 - а - 1 ) / (( а - 1 )( а + 3 ))
2) 2а^2 - а - 1
D = 1 + 8 = 9 = 3^2
a1 = ( 1 + 3 ) : 4 = 1
a2 = ( 1 - 3 ) : 4 = - 0,5
3) ( ( a - 1 )( a + 0,5 )) /(( a - 1 )( a + 3 )) = ( a + 0,5 ) / ( a + 3 )
4) ( ( a + 0,5 ) / ( a + 3 )) : (( 2a + 1 ) / ( a + 3 )) = ( a + 0,5 ) / ( 2a + 1 ) = ( a + 0,5 ) / ( 2( a + 0,5 )) = 1/2 = 0,5
ответ 0,5

MOGZ ответил

5+3 б
1.sin10+cos20

2.sin20+cos10

3.cos20-sin10

4.cos10-sin20